Problem: Let N be a positive even number.

We have a permutation of (1, 2, ..., N), p = (p_1, p_2, ..., p_N). Snuke is constructing another permutation of (1, 2, ..., N), q, following the procedure below.

First, let q be an empty sequence. Then, perform the following operation until p becomes empty:

* Select two adjacent elements in p, and call them x and y in order. Remove x and y from p (reducing the length of p by 2), and insert x and y, preserving the original order, at the beginning of q.



When p becomes empty, q will be a permutation of (1, 2, ..., N).

Find the lexicographically smallest permutation that can be obtained as q.

Constraints

* N is an even number.
* 2 ≤ N ≤ 2 × 10^5
* p is a permutation of (1, 2, ..., N).

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the lexicographically smallest permutation, with spaces in between.

Examples

Input

4
3 2 4 1


Output

3 1 2 4


Input

2
1 2


Output

1 2


Input

8
4 6 3 2 8 5 7 1


Output

3 1 2 7 4 6 8 5

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
#define rint register int
using namespace std;
const int N=2e5+7;
int n,p[N],f[2][N][20],Log[N],pos[N]; 
inline int ask(int opt,int l,int r)
{
	int k=Log[r-l+1];
	return min(f[opt][l][k],f[opt][r-(1<<k)+1][k]);
}
inline pair<int,int> cal(int l,int r)
{
	int x,y;
	x=ask(l&1,l,r);
	y=ask((l-1)&1,pos[x]+1,r);
	return make_pair(-x,-y);
}
priority_queue<pair<pair<int,int>,pair<int,int > > > q;
int main()
{
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
  scanf("%d",&n);
  for(int i=1;i<=n;i++) 
  {
  	scanf("%d",&p[i]);
  	f[i&1][i][0]=p[i];
  	f[(i-1)&1][i][0]=1e9+7;
  	pos[p[i]]=i;
  }
  Log[1]=0;
  for(int i=2;i<=N-7;i++)
    Log[i]=Log[i>>1]+1;
  for(int k=0;k<=1;k++)//奇or偶
    for(int j=1;j<=19;j++)
	  for(int i=1;i+(1<<j)-1<=n;i++)
	    f[k][i][j]=min(f[k][i][j-1],f[k][i+(1<<(j-1))][j-1]);
  q.push({{cal(1,n)},{1,n}}); 
  while(q.size())
  {
  	pair<pair<int,int>,pair<int,int> > tmp=q.top(); q.pop(); //cout<<tmp.second.first<<endl;
  	int x=tmp.first.first,y=tmp.first.second; 
  	printf("%d %d ",-x,-y);
  	x=pos[-x],y=pos[-y];
  	int l=tmp.second.first,r=tmp.second.second; 
  	//cout<<l<<" "<<r<<" "<<x<<" "<<y<<endl;
  	if(l+1<x) q.push({{cal(l,x-1)},{l,x-1}});
  	if(x+1<y-1) q.push({{cal(x+1,y-1)},{x+1,y-1}});
  	if(y+1<r) q.push({{cal(y+1,r)},{y+1,r}});
  }
  return 0;
}